DMSO is an ambidentate ligand that can bind in a κO fashion via the oxygen lone pair and in a κS fashion via the sulfur lone pair.

I know that DMSO can behave as a π-acceptor when it bonds a metal at the sulfur, but what is the symmetry of the $\ce{O\bond{->}M}$ MO when DMSO binds at the oxygen? I have some diagrams of DMSO's HOMO and LUMO MOs, and it seems clear that DMSO-κS is a π-acceptor due to the LUMO's potential to overlap with a $\mathrm{t_{2g}}$-type d orbital, but the HOMO looks like it would do a π interaction. This confuses me because I know from IR that the $\ce{S-O}$ bond is weakened when DMSO binds at the oxygen, and there are some lower energy occupied MOs that would have σ symmetry overlapping with $\mathrm{d}_{z^2}$, for instance.


Is DMSO-κO a σ-donor, a π-donor, or both?

(keep in mind that there are some more σ-donor-looking MOs at energy levels below the HOMO)

Also; The reference below (1) says that DMSO's tendency to bind as DMSO-κS is dependent on the metal having a high enough "electron charge density" to have a sufficient back donation contribution to the $\ce{M\bond{->}S}$ π bond. What does this mean? The paper also says that ruthenium and rhodium prefers to bind as DMSO-κS. Based on Pearson's HSAB concept, I would expect that ruthenium would prefer DMSO-κS because it is a soft acid as a result of its larger second-row ionic radius giving it lower charge density (and S is softer than O), but the paper seems to be saying the opposite. What do they mean by "electron charge density"?

(1) Panina, N. S; Calligaris, M. Inorg. Chim. Acta 2002, 334, 165-171.

  • $\begingroup$ I did some more research and I've noticed from some crystal structures that DMSO-o binds in a nonlinear fashion, so probably only one phase of the HOMO lobe is interacting with a d orbital, and this would make it a σ-donor. If it was a π-donor, it would bind linearly (like with dxz). This is confusing, though, because depopulating this antibonding HOMO should increase S-O bond order, and DMSO-o has decreased IR freq., so there must be some involvement of the lower energy occupied MOs (which are more bonding with respect to the S-O bond). $\endgroup$
    – gannex
    Nov 8, 2016 at 15:58

1 Answer 1


The first thing to remember is that a σ-symmetric bond is always better than a π-symmetric one. Orbitals that can bond with each other are typically oriented in one direction or another (except for s-orbitals, but those can only participate in σ bonds, anyway). Approaching these orbitals from the direction ‘their lobes point towards’ and in the direction ‘your own lobe is pointing towards’ results in a much better overlap than if you attempted to create a π type interaction. Thus, typically any σ interaction will be preferable to any π interaction between a given set of two bonding partners.

To be perfectly honest, I am not aware of any ligand which does not initially bond as a σ donor.

The question remains what will happen if we add π effects into the picture. π effects can either be of the π-Lewis acidic type (e.g. $\ce{CO}$) or of the π-Lewis basic type (e.g. $\ce{F-}$). In the case of DMSO, the LUMO has much greater contributions on sulfur than on oxygen — in line with the general expectation that the LUMO is centred on electropositive partners while HOMOs centre more around the electronegative partners.

Also in line with what oxygen does in other ligands, I would expect DMSO-κO to be a π base, due to the probability of another populated orbital having significant contributions on oxygen. Contrarily, the picture of the LUMO shows how well sulfur in DMSO-κS can act as a Lewis acid.

For Lewis acidic bonding, i.e. π backbonding (also written as M$\leftrightarrows$L), it is beneficial if there is significant charge density on the central metal, i.e. if the metal has a somewhat Lewis basic character. Often, this will be enhanced by the presence of electrons in d-orbitals of the $\mathrm{t_{2g}}$ type.

For Lewis basic ligand bonding, i.e. π forward bonding (also written as L$\rightrightarrows$M), it is beneficial for the metal to be in a Lewis acidic, electron-poor state — often with empty d-orbitals.

In the DMSO-κS context, remember that sulfur carries a partial positive charge and is thus already electron deficient somewhat ($\pm 0$ oxidation state, but it almost has the same electronegativity as carbon, so it is not far away from $\mathrm{+II}$). Therefore, the σ forward bond will be weak per se. Only if sufficient stabilisation through backbonding can be supplied will a DMSO-κS complex be sufficiently stable.

  • $\begingroup$ so you think that DMSO-o is π-basic, then? What about my observation that DMSO-o binds with a M-S-O angle of <180˚ while DMSO-s binds with M-S-O=180˚? I would think that the π-donating interaction from my HOMO would have to bind to a t2g d orbital in a linear fashion, so the nonlinear bonding seems to indicate that it might be more σ-basic. $\endgroup$
    – gannex
    Nov 8, 2016 at 17:23
  • $\begingroup$ @gannex Note that it does not have to be the HOMO doing the sigma donation. I’m also having a hard time imagining the kappa-S version having a $180^\circ \angle(\ce{M-S-O})$ — that would imply that the methyl groups be dangerously close to the metal! $\endgroup$
    – Jan
    Nov 8, 2016 at 17:25
  • $\begingroup$ Oop, you're right. The angle is indeed less than 180˚ for both. I was thinking of the wrong MO. That makes sense given the LUMO I showed... what wouldn't make sense would be a nonlinear bond with DMSO-o given the HOMO I showed IF it was a π-donor interaction. If it indeed does not have to be the HOMO doing the sigma donation, then things make a lot more sense... but I would've thought it would have to be the HOMO. I have a few lower E OMO's that would fit with the observed IR frequencies and bond angles, and I will upload a picture later on... $\endgroup$
    – gannex
    Nov 8, 2016 at 17:53
  • $\begingroup$ @gannex Yes, it doesn’t have to be the HOMO that’s doing the dative bonding but it often is. It just has to be an orbital that is reasonably centred on the donating atom and oriented towards the accepting one. $\endgroup$
    – Jan
    Nov 8, 2016 at 17:55
  • $\begingroup$ OK well if it doesn't have to be the HOMO then theory is looking a lot more in line with observations :-) $\endgroup$
    – gannex
    Nov 8, 2016 at 17:56

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